Problem 6 - Basic Statistics for Business and Economics with Student

Question

A population consists of the following five values: 2,2 , \(4,4,\) and 8 a. List all samples of size \(2,\) and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. c. Compare the dispersion in the population with that of the sample means.

Step-by-Step Solution

Answer

The mean of the sample means (4.11) approximates the population mean (4). The dispersion of the sample means (1.22) is lower than that of the population (2.19).

Step 1: List All Samples

List all possible samples of size 2 from the population \( \{2, 2, 4, 4, 8\} \). The samples are:- \((2, 2)\)- \((2, 4)\)- \((2, 4)\)- \((2, 8)\)- \((4, 2)\)- \((4, 4)\)- \((4, 8)\)- \((8, 2)\)- \((8, 4)\)

Step 2: Calculate Sample Means

Calculate the mean for each sample:- For \((2, 2)\), Mean = \( \frac{2 + 2}{2} = 2\)- For \((2, 4)\), Mean = \( \frac{2 + 4}{2} = 3\)- For \((2, 4)\), Mean = \( \frac{2 + 4}{2} = 3\) (second occurrence)- For \((2, 8)\), Mean = \( \frac{2 + 8}{2} = 5\)- For \((4, 2)\), Mean = \( \frac{4 + 2}{2} = 3\)- For \((4, 4)\), Mean = \( \frac{4 + 4}{2} = 4\)- For \((4, 8)\), Mean = \( \frac{4 + 8}{2} = 6\)- For \((8, 2)\), Mean = \( \frac{8 + 2}{2} = 5\)- For \((8, 4)\), Mean = \( \frac{8 + 4}{2} = 6\)

Step 3: Compute Mean of Sample Means

The means of all samples are \(2, 3, 3, 5, 3, 4, 6, 5, 6\). Calculate the mean of these sample means:Mean of Sample Means = \( \frac{2 + 3 + 3 + 5 + 3 + 4 + 6 + 5 + 6}{9} = \frac{37}{9} \approx 4.11 \).

Step 4: Compute Population Mean

Compute the mean of the population \( \{2, 2, 4, 4, 8\} \):Population Mean = \( \frac{2 + 2 + 4 + 4 + 8}{5} = \frac{20}{5} = 4\).

Step 5: Compare Means

The population mean is \(4\), while the mean of the sample means is approximately \(4.11\). These values are very close, as expected due to the law of large numbers.

Step 6: Compute Population Dispersion

Calculate the standard deviation for the population. The variance is:\[ \frac{(2-4)^2 + (2-4)^2 + (4-4)^2 + (4-4)^2 + (8-4)^2}{5} = \frac{4 + 4 + 0 + 0 + 16}{5} = \frac{24}{5} = 4.8 \].Therefore, the population standard deviation is \( \sqrt{4.8} \approx 2.19 \).

Step 7: Compute Dispersion of Sample Means

Calculate the standard deviation of the sample means \(2, 3, 3, 5, 3, 4, 6, 5, 6\). The variance is:\[ \frac{(2-4.11)^2 + (3-4.11)^2 + (3-4.11)^2 + (5-4.11)^2 + (3-4.11)^2 + (4-4.11)^2 + (6-4.11)^2 + (5-4.11)^2 + (6-4.11)^2}{9} \].After calculation, this results in a variance of approximately \(1.48\).Thus, the standard deviation is \( \sqrt{1.48} \approx 1.22 \).

Step 8: Compare Dispersions

The population standard deviation is approximately \(2.19\), and the standard deviation of the sample means is approximately \(1.22\). The dispersion among the sample means is lower compared to the population dispersion, which is typical as sample means tend to cluster closer around the true mean.

Key Concepts

Population Mean

The population mean is a foundational concept in statistics. It represents the average of all values within a set, or population. Calculating it involves summing all the values and dividing by the number of values.

In our example, the population set is \( \{2, 2, 4, 4, 8\} \). To find the population mean, you sum the values and then divide by the total number of values. So, you get \( \frac{2 + 2 + 4 + 4 + 8}{5} = 4 \).

• Summing the values: \(2 + 2 + 4 + 4 + 8 = 20\)
• Number of values (N): 5
• Population Mean = \( \frac{20}{5} = 4 \)

The result reflects the central tendency of the data, showing where most values cluster.

Sample Dispersion

Sample dispersion indicates how much variation exists in the sample data. One common measure of dispersion is standard deviation, which tells us how much individual sample means tend to deviate from the true mean.

In this exercise, we compute it for the list of sample means: \(2, 3, 3, 5, 3, 4, 6, 5, 6\). The formula for variance, a component of standard deviation, is computed as follows:

• Variance = \( \frac{\sum{(X_i - \bar{X})^2}}{n} \)
• Where \( X_i \) are individual sample means, \( \bar{X} \) is the mean of sample means, and \( n \) is the number of sample means.

The variance calculated results in approximately \(1.48\). Thus, the standard deviation is \(\sqrt{1.48} \approx 1.22\). Lower dispersion among sample means indicates that they tend to be close to the true mean of the population.

Law of Large Numbers

The law of large numbers is a statistical principle that explains why the average of a large number of samples can be expected to approximate the population mean. Basically, as you increase the number of samples taken from a population, the average of the sample means will get closer and closer to the population mean.

In the exercise, the mean of the sample means (approximately \(4.11\)) is indeed very close to the population mean (\(4\)). Although it isn't exactly equal, it illustrates how sample averages converge to the population mean with larger sample sizes.

This principle helps us understand why in research, larger sample sizes generally produce more reliable estimates of the population parameters, since the effects of random variation diminish as sample size grows.

Standard Deviation

Standard deviation is a measure of how spread out numbers are. It is crucial in understanding dispersion in any set of data, whether it's a sample or an entire population.
In simpler terms, standard deviation gives insight into how much the data values differ from the average value.

For a population, you calculate the standard deviation by first finding the variance and then taking the square root. When calculating from the original dataset \( \{2, 2, 4, 4, 8\} \), the variance is \( \frac{24}{5} = 4.8 \), leading to a standard deviation of approximately \(\sqrt{4.8} \approx 2.19\).

• Variance shows how much the numbers differ either above or below the mean.
• Higher standard deviation indicates more variation in values.

In conclusion, standard deviation provides valuable information about data variability and reliability.